how much retardation should be produced to stop a truck moving with velocity of 72km/hr in 5 seconds ?calculate the distance covered by the truck within that time.
Answers
Answer :
The retardation produced is -4 m/s² and the distance covered by truck within that time is 50m
Step-by-step Explanations :
Given : Initial velocity (u) = 72 km/hr
Final velocity (v) = 0
Time(t) = 5 sec
To find : Retardation produced (a) = ?
distance covered by truck within that time = ?
u = 72 km/hr ...(given)
= 72 × 1000/60 × 60
= 72000/3600
= 20 m/s
We know that,.
v = u + at
Substituting the given value in above formula we get,
0 = 20 + a × 5
0 = 20 + 5a
5a = -20.
a = -20/5 = -4 m/s²
Also,
v² = u² + 2as
0 = (20)² + 2 × (-4)s
0 = 400 - 8s
8s = 400
s = 400/8.
s = 50 m
Hence the retardation produced is -4 m/s² and the distance covered by truck within that time is 50m